Given:ABCD is a rhombus.
To prove:DE congruent to BE.
In rombus, we know opposite angle are equal.
so, angle DCB = angle BAD
SINCE, ANGLE DCB= BAD
SO, In triangle DCA
angle DCA=angle DAC
similarly, In triangle ABC
angle BAC=angle BCA
since angle BCD=angle BAD
Therefore, angle DAC =angle CAB
so, opposite sides of equal angle are always equal.
so,sides DC=BC
Now, In triangle DEC and in triangle BEC
1. .DC=BC (from above)............(S)
2ANGLE CED=ANGLE CEB (DC=BC)....(A)
3.CE=CE (common sides)(S)
Therefore,DE is congruent to BE (from S.A.S axiom)
Answer:
x = 5.75
Step-by-step explanation:
4^(x+7) = 8^(2x-3)
But; 4^(x+7) = 2^2(x+7)
8^(2x-3) = 2^3(2x-3)
2^2(x+7) = 2^3(2x-3)
Since the bases are the same;
2(x+7) = 3(2x-3)
2x + 14 = 6x -9
14 + 9 = 6x - 2x
23 = 4x
x = 23/4
<u>x = 5.75</u>
Answer:
7 1/15
Step-by-step explanation:
3 2/5 + 3 4/6 = 7 1/15
